Question

Does the polynomial a4 + 4a2 + 5 have real zeroes?

Answer 1

Answer:

$\huge\bold{\textbf{\textsf{{\color{Black}{αɴѕωєя}}}}}$

In the aforementioned polynomial, let a2 = x.

Now, the polynomial becomes,

x2 + 4x + 5

Comparing with ax2 + bx + c,

Here, b2 – 4ac = 42 – 4(1)(5) = 16 – 20 = -4

So, D = b2 – 4ac < 0

As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.

Answer 2

Answer:

Assertion:

Let p(x)=x4+4x2+5

To finD zeroes of p(x), we consider

p(x)=0

⇒x4+4x2+5=0

⇒x4+4x2+4+1=0

⇒(x2+1)2=−1

Which is not possible. Therefore, the given polynomial has no zero.

Reason:

It is true my remainder theorem.

Hence option D is correct.